Question

A rod of length l rotates with a uniform angular velocity ω about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the rod is
(a) zero
(b) $\frac{1}{2}Bl{\mathrm{\omega }}^2$
(c) Blω²
(d) 2Blω².

Answer 1

(a) zero


Let us consider a small element dx at a distance x from the centre of the rod rotating with angular velocity ω about its perpendicular bisector. The emf induced in the small element of the rod because of its motion is given by
$de=B\omega xdx$
The emf induced between the centre of the rod and one of its end is given by

$$\begin{gathered} \int de={\int }_0^{l}B\omega xdx \\ \Rightarrow e=B\omega {\left[\frac{x^2}{2}\right]}_0^{l/2} \\ \Rightarrow e=\frac{1}{8}B\omega l^2 \end{gathered}$$

The emf at both ends is the same. So, the potential difference between the two ends is zero.